Sampling hw #4


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Funmi Baruwa
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Sampling hw #4

Postby Funmi Baruwa » Tue Jan 12, 2021 11:53 am

Does anyone know how to solve sampling hw #4:

At a certain temperature, the given reaction has an equilibrium constant of Kp=325 .

PCl3(g)+Cl2(g)↽−−⇀PCl5(g)

PCl5 is placed in a sealed container at an initial pressure of 0.0400 bar . What is the total pressure at equilibrium?

what do we do because the initial pressure of the reactants are not given?

Bella Wachter 1A
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Re: Sampling hw #4

Postby Bella Wachter 1A » Tue Jan 12, 2021 11:56 am

We can use the reverse reaction! Since we are starting with only product, we can flip the equation around so that PCl5 is a reactant (and the other two compounds are products). The Kp value of this reverse reaction would be 1/Kp (where Kp is the Kp of the forward reaction).

Then proceed as normal, using an ice table and mole ratios to calculate your partial pressures at equilibrium. Then add them all up at the end to find the total pressure.

Hope this helps :)

Funmi Baruwa
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Joined: Wed Sep 30, 2020 9:50 pm

Re: Sampling hw #4

Postby Funmi Baruwa » Tue Jan 12, 2021 11:57 am

ohhh okay thank you!!1

Cristian Cortes 1L
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Re: Sampling hw #4

Postby Cristian Cortes 1L » Tue Jan 12, 2021 7:18 pm

Bella Wachter 1A wrote:We can use the reverse reaction! Since we are starting with only product, we can flip the equation around so that PCl5 is a reactant (and the other two compounds are products). The Kp value of this reverse reaction would be 1/Kp (where Kp is the Kp of the forward reaction).

Then proceed as normal, using an ice table and mole ratios to calculate your partial pressures at equilibrium. Then add them all up at the end to find the total pressure.

Hope this helps :)


Hi, thank you. I was wondering if the Kp will be 1/355 ? I'm stuck at the end with 1/355=x^2/(.0510-x). Is this correct ?

Bella Wachter 1A
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Re: Sampling hw #4

Postby Bella Wachter 1A » Tue Jan 12, 2021 8:56 pm

Cristian Cortes 1L wrote:
Bella Wachter 1A wrote:We can use the reverse reaction! Since we are starting with only product, we can flip the equation around so that PCl5 is a reactant (and the other two compounds are products). The Kp value of this reverse reaction would be 1/Kp (where Kp is the Kp of the forward reaction).

Then proceed as normal, using an ice table and mole ratios to calculate your partial pressures at equilibrium. Then add them all up at the end to find the total pressure.

Hope this helps :)


Hi, thank you. I was wondering if the Kp will be 1/355 ? I'm stuck at the end with 1/355=x^2/(.0510-x). Is this correct ?



If you were given 355 as the Kp of your forward reaction, then the Kp of the reverse reaction would indeed be 1/355. I'm not sure what your initial value was (we're all given different numbers), but if your initial partial pressure for PCl5 was 0.0510 bar, that setup should work to find the change in partial pressures (x). From there you'd find the equilibrium partial pressures and add them up to get the total pressure.

Tanya Nguyen 1B
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Joined: Wed Sep 30, 2020 9:33 pm

Re: Sampling hw #4

Postby Tanya Nguyen 1B » Tue Jan 12, 2021 9:16 pm

Cristian Cortes 1L wrote:
Bella Wachter 1A wrote:We can use the reverse reaction! Since we are starting with only product, we can flip the equation around so that PCl5 is a reactant (and the other two compounds are products). The Kp value of this reverse reaction would be 1/Kp (where Kp is the Kp of the forward reaction).

Then proceed as normal, using an ice table and mole ratios to calculate your partial pressures at equilibrium. Then add them all up at the end to find the total pressure.

Hope this helps :)


Hi, thank you. I was wondering if the Kp will be 1/355 ? I'm stuck at the end with 1/355=x^2/(.0510-x). Is this correct ?

That is correct, then you would plug it into the quadratic formula to solve for x. You would get two answers one positive and one negative, you would use the positive value as you don't want the equilibrium concentration to have a negative value.

Gina Spagarino 3G
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Re: Sampling hw #4

Postby Gina Spagarino 3G » Wed Jan 13, 2021 10:28 am

Instead of using the reverse reaction's Kp, I just said the equilibrium concentrations of the reactants were equal to x;it still works out mathematically to get the correct answer because the reverse's Kp is the inverse, and the concentrations would be set up inversely as well... so keeping it with the forward Kp and Kp set up give the same results without the fractions for the Kp as long as you are consistent I believe

America Alvarado
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Re: Sampling hw #4

Postby America Alvarado » Wed Jan 13, 2021 11:58 am

Is this why on the sapling feedback the ice box shows reactants asa +x? I have been trying to solve it for the longest and still don't understand. I keep getting imaginary numbers as my x.

Darren1j
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Re: Sampling hw #4

Postby Darren1j » Wed Jan 13, 2021 8:23 pm

For the longest time I kept getting it wrong because I didn't realize total pressure meant add all of the new equilibrium values.

Nan_Guan_1L
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Re: Sampling hw #4

Postby Nan_Guan_1L » Sun Jan 17, 2021 1:53 am

just to add on to the discussion, you can refer to the professor's lecture 2 for guidance. he gave a very similar example during class i remember.

Kaleb Tuliau 3E
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Re: Sampling hw #4

Postby Kaleb Tuliau 3E » Sun Jan 17, 2021 2:03 am

So total pressure is all of our equilibrium pressures combined?

Uyen Trinh 3C
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Re: Sampling hw #4

Postby Uyen Trinh 3C » Sun Jan 17, 2021 2:25 am

I set up an ICE table and then worked it out from there.

Image

Emma Strassner 1J
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Re: Sampling hw #4

Postby Emma Strassner 1J » Sun Jan 17, 2021 11:00 am

Kaleb Tuliau 3E wrote:So total pressure is all of our equilibrium pressures combined?


Yes, the total pressure is all of the equilibrium pressures (found in the last row of the ICE table) combined!

Shannon Moore 2L
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Joined: Wed Feb 26, 2020 12:17 am

Re: Sampling hw #4

Postby Shannon Moore 2L » Sun Jan 17, 2021 11:47 am

I solved it with an ICE box the same way you would solve a Kc problem.

PCl3 Cl2 PCl5
I 0 0 0.0400
C +x +x -x
E +x +x 0.0400 - x

Then plug this into Kp and set it equal to 325.
Solve for x then add together the final pressures.

sabrina ghalambor 2J
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Joined: Wed Sep 30, 2020 9:35 pm

Re: Sampling hw #4

Postby sabrina ghalambor 2J » Sun Jan 17, 2021 1:22 pm

i just redid this problem after the comments here said to use the inverse Kp, but that kept giving me the wrong answer even though ever other part of my problem was set up similarly to everyone's in the comments. i only got the correct answer once i used 391 instead of 1/391, can anyone explain why that worked?

Andy Hon 3E
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Re: Sampling hw #4

Postby Andy Hon 3E » Tue Jan 19, 2021 3:10 pm

My initial thought was to also switch the equation so that PCl5 is on the left and the rest of the equation on the right to show reactants to products. However, the only way to get the correct answer was the say leave PCl 5 as the product and divide it by reactants.

Charlotte Chen 3B
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Joined: Wed Nov 18, 2020 12:29 am

Re: Sampling hw #4

Postby Charlotte Chen 3B » Sun Jan 31, 2021 11:39 pm

sabrina ghalambor 2J wrote:i just redid this problem after the comments here said to use the inverse Kp, but that kept giving me the wrong answer even though ever other part of my problem was set up similarly to everyone's in the comments. i only got the correct answer once i used 391 instead of 1/391, can anyone explain why that worked?


I'm not sure that you have to use the inverse Kp, I didn't need to - the way I did it was I set up an ICE table, with my initial pressure on the right (under PCl5) and 0 for the initial pressures of PCl3 and Cl2, so the equilibrium expression for PCl3 was x, for Cl2 was x, and for PCl5 was 0.0780-x, since 0.0780 was the initial pressure value they gave me. Then I subbed these values into the Kp expression - my Kp value was 413 instead of 391, so I did 431 = [0.0780-x]/[x][x] ; you can then multiply these values out, after which I got the expression 413x^2 +x - 0.0780=0, then I found the roots of x and eliminated the negative root to get x= 0.0126. To find the total pressure I added the equilibrium expressions for each of the compounds pressures and subbed in x, so total pressure = x+x+(0.0780-x), and in this case I got 0.0906. I don't have your initial pressure value so I can't check what I would get for your question specifically, but if you sub in your values into my working hopefully you should get the right answer!


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